Tag: $XY$

While learning about the pumping lemma, I came across the following question:

Given the language L is $ a^n(0|1)^* $ with $ c_0 \cdot c_1 = n $ , where $ c_0 $ indicates the amount of zeros present, use the pumping lemma to prove that this language is not regular. Examples of valid words in L are 0, 1, a01, aa001, etc…

In regular english: a word with leading as, followed by any amount of either 0 or 1 characters, where the amount of zeros multiplied by the amount of ones needs to match the amount of as.

However, the answer key instead opts to introduce two new variables $ m \geq 0 $ , $ n \gt 0 $ , then splits on $ x = a^m $ , $ y = a^n $ , $ z = a^{p-n-m}0^p1 $ (splitting the sequence of a into three parts). Then, they too use $ i = 2 $ to show that $ xy^2z = a^{m}a^{2n}a^{p-n-m}0^p1 = a^{p+n}0^p1 $ , which is not a member of the language (since n was more than 0).

As far as I can see, my attempt adheres to the $ |xy| \leq p $ condition of the pumping lemma: x is empty and $ |y| = p $ . As such, I was under the impression that my answer was correct.

However, the huge increase in complexity of the answer in the answer key leads me to believe that this is not a valid approach. I have the sneaking suspicion this is because my answer actually does violate the $ |xy| \leq p $ condition.

Is my attempt a correct way to prove that the language is not regular? If not, what misconception/mistake did I make along the way?

Let $ X$ be a real or complex vector space endowed with a metric $ d$ which is not induced by a norm, that is there exist no norm $ \| \cdot \|$ on $ X$ such that $ d(x,y)=\| x-y \|$ . Prove or disprove:

This is throwing me off just because the vector space is endowed with a metric that is not norm induced, and I am not sure what all that implies. With that said, I think I may be doing this problem naively. Anyway, here are my solutions:

Let $ X$ and $ Y$ be random variables such that $ E(|X|), E(|Y|)<\infty$ and $ E(X|Y)=E(Y|X)$ a.s. Then is it true that $ X=Y$ a.s. ? If this is not true in general, what happens if we also assume $ X,Y$ are identically distributed ?

If $ Y=Y_1\cup Y_2$ are such that $ X=Y_1\cap Y_2$ and if each $ Y_i$ is a cone over $ X$ , then $ Y$ is the suspension of $ X$

But in page 90, in the proof of the Proposition 2.5, they use this in a way I don’t understand. They prove that $ \mid X\mid \simeq C\mid E\mid $ and $ \mid Y\mid \simeq C\mid F\mid$ , but those aren’t cones over $ \mid X\cap Y\mid$ , so why I can continue the proof of this proposition saying that this implications make $ \mid G\mid \simeq \sum(\mid X\cap Y\mid)$ ?